16–58.
If the block at C is moving downward at 4 ft/s, determine
the angular velocity of bar AB at the instant shown.
C
3 ft
A
30°
ω AB
SOLUTION
vC = 4 ft/s
B
2 ft
Kinematic Diagram: Since link AB is rotating about fixed point A, then vB is always
directed perpendicular to link AB and its magnitude is vB = vAB rAB = 2vAB. At
the instant shown, vB is directed towards the negative y axis. Also, block C is moving
downward vertically due to the constraint of the guide. Then vc is directed toward
negative y axis.
Velocity Equation: Here, rC>A = {3 cos 30°i + 3 sin 30°j} ft = {2.598i + 1.50j} ft.
Applying Eq. 16–16, we have
vC = vB + vBC * rC>B
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
- 4j = - 2vAB j + (vBCk) * (2.598i + 1.50j)
-4j = - 1.50vBCi + (2.598vBC - 2vAB)j
Equating i and j components gives
0 = - 1.50vBC
- 4 = 2.598(0) - 2vAB
vBC = 0
vAB = 2.00 rad s
Ans.
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16–62.
If the flywheel is rotating with an angular velocity of
vA = 6 rad>s, determine the angular velocity of rod BC at
the instant shown.
1.5 m
A
0.3 m
vA ⫽ 6 rad/s
O
C
B
SOLUTION
60⬚ 0.6 m
D
Rotation About a Fixed Axis: Flywheel A and rod CD rotate about fixed axes,
Figs. a and b. Thus, the velocity of points B and C can be determined from
vB = vA * rB = (- 6k) * (- 0.3j) = [-1.8i] m>s
vC = vCD * rC = (vCDk) * (0.6 cos 60° i + 0.6 sin 60° j)
= - 0.5196vCD i + 0.3vCD j
vB = vC + vBC * rB>C
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
General Plane Motion: By referring to the kinematic diagram of link BC shown in
Fig. c and applying the relative velocity equation, we have
- 1.8i = - 0.5196vCD i + 0.3vCD j + (vBC k) * (- 1.5i)
-1.8i = - 0.5196vCD i + (0.3vCD - 1.5vBC)j
Equating the i and j components
- 1.8 = - 0.5196vCD
0 = 0.3vCD - 1.5vBC
Solving,
vCD = 3.46 rad>s
vBC = 0.693 rad>s
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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16–111.
Crank AB rotates with the angular velocity and angular
acceleration shown. Determine the acceleration of the
slider block C at the instant shown.
0.4 m
B
vAB ⫽ 4 rad/s
aAB ⫽ 2 rad/s2
A
30⬚
0.4 m
30⬚
C
SOLUTION
Angular Velocity: Since crank AB rotates about a fixed axis, Fig. a,
vB = vAB rB = 4(0.4) = 1.6 m>s
The location of the IC for link BC is indicated in Fig. b. From the geometry of this
figure,
rB>IC = 0.4 m
Then
vB
1.6
=
= 4 rad>s
rB>IC
0.4
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
vBC =
Acceleration and Angular Acceleration: Since crank AB rotates about a fixed axis,
Fig. a
aB = aAB * rB - vAB2rB
= (- 2k) * (0.4 cos 30°i + 0.4 sin 30°j) - 4 2(0.4 cos 30°i + 0.4 sin 30°j)
= [ -5.143i - 3.893j] m>s2
Using this result and applying the relative acceleration equation by referring to Fig. c,
aC = aB + aBC * rC>B - vBC2 rC>B
aCi = (-5.143i - 3.893j) + (aBCk) * (0.4 cos 30°i - 0.4 sin 30°j) - 42(0.4 cos 30°i - 0.4 sin 30°j)
aCi = (0.2aBC - 10.69)i + (0.3464aBC - 0.6928)j
Equating the i and j components, yields
aC = 0.2aBC - 10.69
0 = 0.3464aBC - 0.6928
(1)
(2)
Solving Eqs. (1) and (2),
aBC = 2 rad>s2
aC = - 10.29 m>s2 = 10.3 m>s2 ;
Ans.
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16–119.
The wheel rolls without slipping such that at the instant
shown it has an angular velocity V and angular acceleration
A. Determine the velocity and acceleration of point B on
the rod at this instant.
A
2a
O
V, A
a
B
SOLUTION
vB = vA + vB/A (Pin)
+ v =
;
B
1
Qv 22aR + 2av¿ a b
2
22
+c O = v¿ =
1
1
22
Qv22aR + 2av¿ a
23
b
2
v
23
vB = 1.58 va
a A = aO + aA/O (Pin)
(a A)x + (aA)y = aa + a(a) + v2a
;
T
;
T
:
(a A)x = aa - v2a
(a A)y = aa
a B = aA + aB/A (Pin)
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
Ans.
v 2 23
1
a B = aa - v2a + 2a(a¿) a b - 2a a
b
2
2
23
O = -aa + 2aa¿ a
2
23
b + 2a a
a¿ = 0.577a - 0.1925v2
a B = 1.58aa - 1.77v2a
v
2
1
b a b
2
23
Ans.
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17–34.
The trailer with its load has a mass of 150 kg and a center of
mass at G. If it is subjected to a horizontal force of
P = 600 N, determine the trailer’s acceleration and the
normal force on the pair of wheels at A and at B. The
wheels are free to roll and have negligible mass.
G
P
1.25 m
0.25 m
SOLUTION
0.75 m
Equations of Motion: Writing the force equation of motion along the x axis,
+ ©F = m(a ) ;
:
x
G x
600 = 150a
a = 4 m>s2 :
B
0.25 m
A
600 N
0.5 m
1.25 m
Ans.
Using this result to write the moment equation about point A,
a + ©MA = (Mk)A ;
150(9.81)(1.25) - 600(0.5) - NB(2) = -150(4)(1.25)
NB = 1144.69 N = 1.14 kN
Ans.
Using this result to write the force equation of motion along the y axis,
NA + 1144.69 - 150(9.81) = 150(0)
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
+ c ©Fy = m(a G)y ;
NA = 326.81 N = 327 N
Ans.
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17–39.
A
The uniform bar of mass m is pin connected to the collar,
which slides along the smooth horizontal rod. If the collar is
given a constant acceleration of a, determine the bar’s
inclination angle u. Neglect the collar’s mass.
L
a
u
SOLUTION
Equations of Motion: Writing the moment equation of motion about point A,
+ ©MA = (Mk)A;
mg sin u a
Ans.
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
a
u = tan-1 a b
g
L
L
b = ma cos u a b
2
2
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17–43.
1 ft
Determine the acceleration of the 150-lb cabinet and the
normal reaction under the legs A and B if P = 35 lb. The
coefficients of static and kinetic friction between the
cabinet and the plane are ms = 0.2 and mk = 0.15,
respectively. The cabinet’s center of gravity is located at G.
P
G
4 ft
SOLUTION
Equations of Equilibrium: The free-body diagram of the cabinet under the static
condition is shown in Fig. a, where P is the unknown minimum force needed to move
the cabinet. We will assume that the cabinet slides before it tips. Then,
FA = msNA = 0.2NA and FB = msNB = 0.2NB.
+ ©F = 0;
:
x
P - 0.2NA - 0.2NB = 0
(1)
+ c ©Fy = 0;
NA + NB - 150 = 0
(2)
+ ©MA = 0;
NB(2) - 150(1) - P(4) = 0
(3)
1 ft
3.5 ft
A
B
Solving Eqs. (1), (2), and (3) yields
NA = 15 lb
NB = 135 lb
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
P = 30 lb
Since P 6 35 lb and NA is positive, the cabinet will slide.
Equations of Motion: Since the cabinet is in motion, FA = mkNA = 0.15NA and
FB = mkNB = 0.15NB. Referring to the free-body diagram of the cabinet shown in
Fig. b,
+ ©F = m(a ) ;
:
x
G x
35 - 0.15NA - 0.15NB = a
+ ©F = m(a ) ;
:
x
G x
NA + NB - 150 = 0
+ ©MG = 0;
150
ba
32.2
NB(1) - 0.15NB(3.5) - 0.15NA(3.5) - NA(1) - 35(0.5) = 0
Solving Eqs. (4), (5), and (6) yields
a = 2.68 ft>s2
NA = 26.9 lb
(4)
(5)
(6)
Ans.
NB = 123 lb
Ans.
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